On Mar 6, 2015, at 6:43 AM, Henry Baker <hbaker1@pipeline.com> wrote:
Situation 1: Non-rotating Earth. The child's circular pool (R=~3') is placed on a lazy susan and rotated at 1 revolution per day. Result: concave parabolic surface. This is the canonical Newton telescope, except for the extremely slow rotational rate. Not quite: the surface of the pool would still be acted upon by a spherical gravitational field.
A fluid particle at rest with respect to the rotating frame sees an additional potential energy function -m(wr)^2/2, where m is the particle mass, w is the angular velocity, and r is the distance from the rotation axis. Let h(r) be the small extra height of the pool surface above the round earth. The equilibrium h(r) is the function for which -m(wr)^2/2 +mgh(r) is a constant, whose value is 0 if we take h(0)=0. Therefore h(r) = (wr)^2/(2g). What would h(r) have to be for the pool surface to be a flat plane? By simple geometry h(r) = r^2/(2R) where R>>r is the radius of the earth. So if you want to make a perfectly flat mirror the angular velocity would have to be w=sqrt(g/R) Comparing with the earth’s angular velocity w_E we get w/w_E = 17.05. I think I’ll use this on my next mechanics prelim!
Situation 2: Rotating Earth; child's pool located at the South Pole. Pool still rotates at 1 revolution per day. Pool & water within pool can't tell the difference between situation 1 and situation 2. Result: same concave parabolic surface.
If you rotate a child's pool, the water will rush towards the edges; the only thing keeping the water from flowing out is the pool wall, hence the pool wall will feel not only the static load of the water, but also an incremental force due to the centrifugal force of the rotating water within the pool.
Thus, it is the _constraint_ of the pool wall which keeps the surface of the pool water in a more-or-less parabolic shape.